Calculate the circumference/diameter in this relativistic problem

[LCSphysicist]

Homework Statement

> A record turntable of radius $R$ rotates at angular velocity $w$ (Fig.
> 12.15). The circumference is presumably Lorentz-contracted, but the radius (being perpendicular to the velocity) is not. What's the ratio
> of the circumference to the diameter, in terms of $w$ and $R$?

Relevant Equations

.

$$ds^2 = dt^2 - r^2 d\theta^2 = d\theta^2((dt/d\theta)^2-r^2) = d\theta^2((1/w)^2-r^2)$$ $$C = \Delta S = \int ds = \int \sqrt{(1/w^2-r^2)}d\theta = 2\pi \sqrt{(1/w^2-r^2)} $$ $$C/2r = \frac{\pi \sqrt{(1/w^2-r^2)}}{r} = \pi \sqrt{\frac{1}{(wr)^2}-1}$$

But this answer is wrong. And i don't know why. I can't see the problem on my reasoning. I have just calculated the distance at the rotating reference frame.

 

[Orodruin]

I have just calculated the distance at the rotating reference frame.

No, you have not. You have computed the proper time taken for an observer situated at the edge to complete a full turn and divided it by $2r$.